Explanation Given data: Shortest wavelength is 248 nm. Light wavelength is 414 nm. Formula used: Consider the work function equation. E = h c λ (1) Here, h is the planks constant, which has a standard value as 6.63 × 10 − 34 J ⋅ s , c is the speed of light , which has a standard value as 3 × 10 8 m s , and λ is the wavelength. Consider the expression for wavelength. λ = h c E n + 1 − E n (2) Here, E n is the energy of n level, and E n + 1 is the energy of n + 1 level. Explanation: For shortest wavelength, substitute 6.63 × 10 − 34 J ⋅ s for h , 3 × 10 8 m s for c , and 248 nm for λ in equation (1), E = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s ) 248 nm = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s ) 248 × 10 − 9 m = 0.08 × 10 − 17 J Consider the conversion factor for J to eV. 1 J = 1 1.6 × 10 − 19 eV Energy in terms of eV is given as, E = 0.08 × 10 − 17 1.6 × 10 − 19 eV = 0 .0501 × 10 2 eV = 5.01 eV For light wavelength, substitute 6.63 × 10 − 34 J ⋅ s for h , 3 × 10 8 m s for c , and 414 nm for λ in equation (1), E = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s ) 414 nm = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s ) 414 × 10 − 9 m = 0

College Physics: A Strategic Approach (4th Edition)

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ISBN: 9780134609034

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 28, Problem 49P

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The third wavelength of the energy spectrum.

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Chapter 28, Problem 48P

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The third wavelength of the energy spectrum.

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The third wavelength of the energy spectrum.

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Chapter 28 Solutions